Simplify the following expression: $y = \dfrac{3x^2- 23x+40}{3x - 8}$
Explanation: First use factoring by grouping to factor the expression in the numerator. This expression is in the form ${A}x^2 + {B}x + {C}$ First, find two values, $a$ and $b$ , so: $ \begin{eqnarray} {ab} &=& {A}{C} \\ {a} + {b} &=& {B} \end{eqnarray} $ In this case: $ \begin{eqnarray} {ab} &=& {(3)}{(40)} &=& 120 \\ {a} + {b} &=& &=& {-23} \end{eqnarray} $ In order to find ${a}$ and ${b}$ , list out the factors of $120$ and add them together. The factors that add up to ${-23}$ will be your ${a}$ and ${b}$ When ${a}$ is ${-8}$ and ${b}$ is ${-15}$ $ \begin{eqnarray} {ab} &=& ({-8})({-15}) &=& 120 \\ {a} + {b} &=& {-8} + {-15} &=& -23 \end{eqnarray} $ Next, rewrite the expression as $({A}x^2 + {a}x) + ({b}x + {C})$ $ ({3}x^2 {-8}x) + ({-15}x +{40}) $ Factor out the common factors: $ x(3x - 8) - 5(3x - 8)$ Now factor out $(3x - 8)$ $ (3x - 8)(x - 5)$ The original expression can therefore be written: $ \dfrac{(3x - 8)(x - 5)}{3x - 8}$ We are dividing by $3x - 8$ , so $3x - 8 \neq 0$ Therefore, $x \neq \frac{8}{3}$ This leaves us with $x - 5; x \neq \frac{8}{3}$.